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I am a novice who purchased several TO-220 L4940v5 chips thinking they were drop-in replacements for the inefficient TO-220 7805s that I've been using. But not a single one seems to work. No heat, no voltage output, nothing. Thinking I had the pins backward I tried a fresh one with the in & out swapped, but still no result.

Is the L4940v5 not pin-compatible with the 7805?

EDIT: Here is what I built. It is the same as the reference diagram on the L4904v5 datasheet, except my output capacitor is larger (it's what I had). Diagram of test circuit

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    \$\begingroup\$ Please show the exact manner in which you connected the parts, including any decoupling capacitors included in your design. \$\endgroup\$
    – nanofarad
    May 1, 2020 at 22:08
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    \$\begingroup\$ Note that they are a low drop-out version. That just means they'll work on a lower input voltage. They're just as inefficient because they are linear regulators. \$\endgroup\$
    – Transistor
    May 1, 2020 at 22:16
  • \$\begingroup\$ @Transistor I extract from your response that the term "efficient" only refers to the ratio of output power to input power, rather than the total amount of usable energy I can extract from the batteries? Having a LDO device lets me run on fewer batteries, and run them to a lower voltage. That means I extract more total joules from the batteries before my circuit fails. Is there a term for that? \$\endgroup\$
    – Moby Disk
    May 2, 2020 at 16:57
  • \$\begingroup\$ What battery were you intending to use? \$\endgroup\$
    – Bruce Abbott
    May 2, 2020 at 17:27
  • \$\begingroup\$ @BruceAbbott I'm using AA and AAA alkalines. My prior circuits used 6x1.5 + 7805. I bought these to try 5x1.5 + L4940. I think after having done this its time for me to switch to something like LiPo + boost converters. \$\endgroup\$
    – Moby Disk
    May 2, 2020 at 23:21

2 Answers 2

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Getting no output sounds like something is wrong, like maybe the parts are defective. Hopefully you purchased them from a reliable source.

They're not exact replacements because they can withstand less input voltage (30V rather than 35V) and they draw rather nasty amounts of Iq (as much as 100mA typical) when the output current is high and the regulator is dropping out (this hideous behavior is typical of such antediluvian regulators made with low-hFE lateral PNP pass transistors). I suppose it's possible this is causing your source to current limit and restart, resulting in an endless loop.

They don't say so in the datasheet but the regulator is probably not guaranteed stable with ceramic output caps, and needs a fat-ish 22uF or more aluminum electrolytic or tantalum cap with relatively high ESR on the output. The LM7805 is much more forgiving.

There's no power dissipation advantage in dropping one of these in place of a 7805, not at all, under some conditions just the opposite. There are switching regulators that you can use such as the Murata OKI-78SR-5/1.5-W36-C which actually are very efficient and are (more-or-less) drop-in replacements allowing you to leave off the heatsink. Be sure to read the datasheet.

enter image description here

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    \$\begingroup\$ You are correct that the part was defective. I am new enough to this that when something doesn't work I assume it is my fault, not the part. I didn't want to test the others until I had my circuit validated. But with your advice I proceeded to test the remaining ones. I bought a lot of 10 from either aliexpress or ebay. After testing the lot, 4 of them don't work and 6 do. Thank you! \$\endgroup\$
    – Moby Disk
    May 2, 2020 at 16:54
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    \$\begingroup\$ @MobyDisk You're welcome. I hope you learned something about taking extreme care when buying from unreliable sources. A defective voltage regulator can destroy everything connected to it if it fails 'on'. \$\endgroup\$
    – Spehro Pefhany
    May 2, 2020 at 16:55
  • \$\begingroup\$ Also, thanks for the suggestion for the alternative part and the information about the current draw. \$\endgroup\$
    – Moby Disk
    May 2, 2020 at 16:59
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This is a supplementary answer to respond to the comments:

I extract from your response that the term "efficient" only refers to the ratio of output power to input power, rather than the total amount of usable energy I can extract from the batteries? Having a LDO device lets me run on fewer batteries, and run them to a lower voltage. That means I extract more total joules from the batteries before my circuit fails.

Yes, you are correct that with an LDO regulator you can run at lower battery voltage. There may be an odd catch if you use a higher voltage with a linear regulator. Let's do some rough maths.

  • Cells are 1000 mAh, 1.5 V. This means they have 1.5 Wh energy capacity per cell. We have a choice of running at 6 V or 7.5 V.

6 V option

  • Vbatt = 6 V, I = 100 mA.
  • Power = Vi = 0.6 W.
  • Run time = Ebatt / Pload = 4 × 1.5 / 0.6 = 10 hours.

7.5 V option

  • Vbatt = 7.5 V, I = 100 mA.
  • Power = Vi = 7.5 W.
  • Run time = Ebatt / Pload = 5 × 1.5 / 0.75 = 10 hours1.

1 You could do better than this by discharging further on the 7.5 V battery.

The point to note here is that you get the same runtime with 5 cells instead of 4 but you've wasted one cell.

Is there a term for that?

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  • \$\begingroup\$ Thank you for taking the time to add this answer. I am completely aligned with your point that any extra voltage just gets burned away by the regulator. At you say in caveat (1), the actual runtime calculation is more complex since one must account for the minimum voltage. Not all the joules from the battery gets used. That's why the 7805 gets less runtime than the 4940. With 4 cells @ 6V, the 7805 gives me 0 minutes of runtime since it needs >2V dropout. Since I have this circuit built, maybe I'll time them with various loads and make a chart showing actual real-world runtime. \$\endgroup\$
    – Moby Disk
    May 2, 2020 at 23:18
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    \$\begingroup\$ Your maths are pretty rough. However by pure coincidence a good AA alkaline will last ~10 hours at 100mA to 1.25V (x4 = 5V). That's about half the capacity to 1.1V. The 5 cell pack should last twice as long. \$\endgroup\$
    – Bruce Abbott
    May 3, 2020 at 2:18

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